Question

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is

S2(g)+C(s)−⇀↽−CS2(g)Kc=9.40 at 900 KS2(g)+C(s)↽−−⇀CS2(g)Kc=9.40 at 900 K

How many grams of CS2(g)CS2(g) can be prepared by heating 16.5 mol S2(g)16.5 mol S2(g) with excess carbon in a 9.40 L9.40 L reaction vessel held at 900 K until equilibrium is attained?

mass of CS2(g)CS2(g):

Answer 1

Ans :-

Initial molar concentration of S₂ = Number of moles of S₂ / Volume in L

= 16.5 mol / 9.40 L

= 1.755 M

ICE table is :

................................S₂(g)........................+........................C(s)<------------------------------> CS₂(g)

Initial .......................1.755 M ..................................................................................................0.0 M

Change.....................-y...............................................................................................................+y

Equilibrium................(1.755 - y) M.............................................................................................y M

Where, y = Moles dissociated per mole

Expression of Equilibrium constant i.e. K_c(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_c = [CS₂]/[S₂]

9.40 = y / (1.755 - y)

9.40 (1.755 - y) = y

y = 16.497 - 9.40 y

10.40 y = 16.497

y = 16.497 / 10.40 = 1.586 M

So, Moles of CS₂ at equilibrium = 1.586 M x 9.40 L = 14.908 mol

Mass of CS₂ = Moles x Gram molar mass = 14.908 mol x 76.139 g/mol = 1135.08 g

Therefore,

Mass of CS₂ obtained = 1135.08 g